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My textbook states that at equilibrium thermodynamic potentials are minimized. I am having trouble understanding how this minimization work and how to visualize it. For example, the Helmholtz free energy of an ideal gas is: $$A(T, V, N)=NK_{B}T(\frac{3}{2}-ln((\frac{{3*K_{B}T}}{2})^{\frac{3}{2}}(\frac{V}{N}))-C)$$

When I plot A vs T, A vs V, or A vs N, I don't see any minima in the graph. So I wanted to know what minimizing (A) at equilibrium exactly means.

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The minimization of thermodynamic potentials does not refer to the minimization of the thermal equation of state as function of its explicit variables. Instead it refers to the result of an interaction of the system with its environment.

For example, imagine you have two systems that are isolated from each other and their common environment both are in equilibrium having $S_1$ and $S_2$ entropies, resp. Now let them interact but stay still isolated from the environment, then the 2nd law says then when they reach common equilibrium with each other then the total entropy $S_0$ will not decrease, i.e, $S_1+S_2 \le S_0$ This is what maximum entropy means in this: equilibrium is achieved at the maximum of the total entropy.

Now take a system that is in equilibrium and isolated form its environment having free energy $A_0$. Now let this system interact only with an entropy ("heat") reservoir at a fixed temperature $T_1$. Wait until the system reaches equilibrium with the reservoir now both at $T_1$ by exchanging some amount of entropy with it. Now the new free energy of the system will not exceed its initial, ie., $A_1 \le A_0$. In other words, the thermal equilibrium of the system with its environment is reached when the free energy is minimum.

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  • $\begingroup$ so the minimization process of A is with respect to the energy exchange with the bath not the natural variables of A? I.e the value of A after the system reaches equilibrium with the bath is found at the value of E where $\frac{\partial A}{\partial E}=0$? Where E is the energy of the system. $\endgroup$
    – Abe
    Commented Nov 19, 2022 at 21:50
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    $\begingroup$ Let the reservoir's energy be $U_1^r$ and let the total energy of the system and reservoir be $U^*=U+U_1^r$, this is constant per the 1st law so $dU^*=0$ and since the interaction is purely thermal $0=dU^*=dU - T_1dS=d(U_1-T_1S) = dA$ where $dS$ is the entropy supplied by the reservoir to the system. In other words in equilibrium, $dA=0$ relative to any small variation (fluctuation) but since in equilibrium the total energy is also minimum so must be $A$ too. Assumed that no work is done, only thermal exchange to settle, the minimum here refers to internal uncontrolled parameters. $\endgroup$
    – hyportnex
    Commented Nov 19, 2022 at 22:42
  • $\begingroup$ "In other words, the thermal equilibrium of the system with its environment is reached when the free energy is minimum." What does this mean if not, "The minimization of thermodynamic potentials does not refer to the minimization of the thermal equation of state as function of its explicit variables?" $\endgroup$
    – Jbag1212
    Commented Nov 20, 2022 at 8:06
  • $\begingroup$ Take a system and place it in a reservoir at $T_1.$ The free energy should decrease. Now take it to a reservoir at $T_2$, the free energy should decrease. Now take it back to the same reservoir again at $T_1$, is the free energy going to decrease again? $\endgroup$
    – Jbag1212
    Commented Nov 20, 2022 at 8:07
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    $\begingroup$ @Jbag imagine liquid and its vapor enclosed in a rigid fixed volume, $V_{tot}=const$ and $N=const$ but the fraction of steam to liquid is variable. That ratio is an internal variable that you do not directly control, it will settle at any given temperature to the minimum total Helmholtz free energy. If the external pressure (and not the total volume) is constrained then the equilibrium ratio will settle at he total minimum Gibbs free energy. $\endgroup$
    – hyportnex
    Commented Nov 20, 2022 at 15:58

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